Integrand size = 18, antiderivative size = 223 \[ \int x^5 \left (a+b x^2+c x^4\right )^p \, dx=-\frac {b (2+p) \left (a+b x^2+c x^4\right )^{1+p}}{4 c^2 (1+p) (3+2 p)}+\frac {x^2 \left (a+b x^2+c x^4\right )^{1+p}}{2 c (3+2 p)}+\frac {2^{-1+p} \left (2 a c-b^2 (2+p)\right ) \left (-\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{\sqrt {b^2-4 a c}}\right )^{-1-p} \left (a+b x^2+c x^4\right )^{1+p} \operatorname {Hypergeometric2F1}\left (-p,1+p,2+p,\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{2 \sqrt {b^2-4 a c}}\right )}{c^2 \sqrt {b^2-4 a c} (1+p) (3+2 p)} \]
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Time = 0.15 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1128, 756, 654, 638} \[ \int x^5 \left (a+b x^2+c x^4\right )^p \, dx=\frac {2^{p-1} \left (2 a c-b^2 (p+2)\right ) \left (a+b x^2+c x^4\right )^{p+1} \left (-\frac {-\sqrt {b^2-4 a c}+b+2 c x^2}{\sqrt {b^2-4 a c}}\right )^{-p-1} \operatorname {Hypergeometric2F1}\left (-p,p+1,p+2,\frac {2 c x^2+b+\sqrt {b^2-4 a c}}{2 \sqrt {b^2-4 a c}}\right )}{c^2 (p+1) (2 p+3) \sqrt {b^2-4 a c}}-\frac {b (p+2) \left (a+b x^2+c x^4\right )^{p+1}}{4 c^2 (p+1) (2 p+3)}+\frac {x^2 \left (a+b x^2+c x^4\right )^{p+1}}{2 c (2 p+3)} \]
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Rule 638
Rule 654
Rule 756
Rule 1128
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int x^2 \left (a+b x+c x^2\right )^p \, dx,x,x^2\right ) \\ & = \frac {x^2 \left (a+b x^2+c x^4\right )^{1+p}}{2 c (3+2 p)}+\frac {\text {Subst}\left (\int (-a-b (2+p) x) \left (a+b x+c x^2\right )^p \, dx,x,x^2\right )}{2 c (3+2 p)} \\ & = -\frac {b (2+p) \left (a+b x^2+c x^4\right )^{1+p}}{4 c^2 (1+p) (3+2 p)}+\frac {x^2 \left (a+b x^2+c x^4\right )^{1+p}}{2 c (3+2 p)}-\frac {\left (2 a c-b^2 (2+p)\right ) \text {Subst}\left (\int \left (a+b x+c x^2\right )^p \, dx,x,x^2\right )}{4 c^2 (3+2 p)} \\ & = -\frac {b (2+p) \left (a+b x^2+c x^4\right )^{1+p}}{4 c^2 (1+p) (3+2 p)}+\frac {x^2 \left (a+b x^2+c x^4\right )^{1+p}}{2 c (3+2 p)}+\frac {2^{-1+p} \left (2 a c-b^2 (2+p)\right ) \left (-\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{\sqrt {b^2-4 a c}}\right )^{-1-p} \left (a+b x^2+c x^4\right )^{1+p} \, _2F_1\left (-p,1+p;2+p;\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{2 \sqrt {b^2-4 a c}}\right )}{c^2 \sqrt {b^2-4 a c} (1+p) (3+2 p)} \\ \end{align*}
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 0.43 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.73 \[ \int x^5 \left (a+b x^2+c x^4\right )^p \, dx=\frac {1}{6} x^6 \left (\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^2+c x^4\right )^p \operatorname {AppellF1}\left (3,-p,-p,4,-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right ) \]
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\[\int x^{5} \left (c \,x^{4}+b \,x^{2}+a \right )^{p}d x\]
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\[ \int x^5 \left (a+b x^2+c x^4\right )^p \, dx=\int { {\left (c x^{4} + b x^{2} + a\right )}^{p} x^{5} \,d x } \]
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\[ \int x^5 \left (a+b x^2+c x^4\right )^p \, dx=\int x^{5} \left (a + b x^{2} + c x^{4}\right )^{p}\, dx \]
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\[ \int x^5 \left (a+b x^2+c x^4\right )^p \, dx=\int { {\left (c x^{4} + b x^{2} + a\right )}^{p} x^{5} \,d x } \]
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\[ \int x^5 \left (a+b x^2+c x^4\right )^p \, dx=\int { {\left (c x^{4} + b x^{2} + a\right )}^{p} x^{5} \,d x } \]
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Timed out. \[ \int x^5 \left (a+b x^2+c x^4\right )^p \, dx=\int x^5\,{\left (c\,x^4+b\,x^2+a\right )}^p \,d x \]
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